Question

A force of 1.200×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 615 N. If he starts from rest and is on a level road, what speed ???? will he be going after 25.0 m?

Answer 1

Answer:

15.45 m/s

Explanation:

F = 1.2 x 10^3 N, Fk = 615 N, u = 0, s = 25 m

Let the speed after traveling 25 m is v.

mass = F / g = 1200 / 9.8 = 122.45 kg

Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N

acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2

Use third equation of motion

v^2 = u^2 + 2 a s

v^2 = 0 + 2 x 4.78 x 25 = 238.88

v = 15.45 m/s